Question

When dividing with the de moivre's, how do I do it?

Do I divide the r values and subtract the theta values?

For example if I have: 14.76(cos(0.49)+isin(0.49))/3(cos(1.43)+isin(1.43))

Answer 1

@satellite73 @jim_thompson5910 @ganeshie8 @mathstudent55 @zepdrix @Nnesha @Luigi0210 @whpalmer4 @sleepyjess @mathmate @KyanTheDoodle @misty1212

Answer 2

Yes.

Have you learned about the exponential form of the complex number?

Answer 3

It sounds familiar, if I see it I will probably remember

Answer 4

That certainly sheds some light on these rules.\[\Large\rm r( \cos \theta + i \sin \theta)= r e^{i \theta}\]So if we have:\[\large\rm \frac{r_1(\cos \alpha+i \sin \alpha)}{r_2(\cos \beta+i \sin \beta)}=\frac{r_1 e^{i \alpha}}{r_2e^{i \beta}}=\frac{r_1}{r_2}e^{i(\alpha-\beta)}\]When you convert to exponential form, you can see that it's just an exponent rule being applied to the angles.

Answer 5

\[\Large\rm \frac{r_1}{r_2}e^{i(\alpha-\beta)}=\frac{r_1}{r_2}(\cos(\alpha-\beta)+i \sin (\alpha-\beta))\]

Answer 6

sure is a damn sight easier than using the "addition angle" formula isn't it? just the laws of exponents is all, a much quicker explanation !

Answer 7

So in this case 14.76/44.29(cis(0.49-1.92))?

Answer 8

Oh wait, nvm

Answer 9

sorry, I added the wrong equation

Answer 10

14.76/3(cis(0.49-1.43))

Answer 11

mmm ya that looks better! :)

Answer 12

So if I simplify this, it is the answer?

Answer 13

(14.76/3) cis(0.49-1.43)
^ Just making sure that you know the cis is not in the denominator with the 3

Yes.
Just make sure you have the correct form, polar or rectangular or whatever :p

Answer 14

4.92(cis(-0.94)) Is a negative acceptable?

Answer 15

Hehe, you tell me :)
If you're supposed to have a positive angle, you could can take a trip around the circle. Add 2pi or 360 degrees to your angle (Depending on whether or not that was in radians or degrees).

Answer 16

But in general, a negative is fine. :o

Answer 17

awesome! Is rectangular form a+bi?

Answer 18

Not for this equation

Answer 19

Just curious

Answer 20

yes, thats what i was referring to.
i guess they call it "standard form" or something though, my bad :)

Answer 21

Actually for this problem they wanted polar. I was curious, because for another problem I am asked for rectangular form.

Answer 22

Also, is the complex conjugate of a r(cos(theta)+isin(theta)) r(cos(theta)-isin(theta))?

Answer 23

mmm yes!

you can actually solve this problem by multiplying top and bottom by the complex conjugate of the denominator. That's another approach to try sometime :) It's kinda neat.

Answer 24

err i guess we shouldn't include the r's when we're talking about complex conjugate.
It's more accurate to say (cos theta- i sin theta) is the conjugate of (cos theta+ i sin theta).
Let's leave the r's out of there :3

Answer 25

ah gotcha

Answer 26

But if there was an r, it would stay the same?

Answer 27

Well it's just that umm...

When you multiply complex conjugates,
you end up with `the sum of squares`.

Example:
\(\Large\rm (\cos x+i\sin x)(\cos x-i\sin x)=\cos^2x+\sin^2x\)
If we threw some r's into the mix though,
\(\Large\rm r(\cos x+i\sin x)\cdot r(\cos x-i\sin x)=r^2(\cos^2x+\sin^2x)\)
We get the sum of squares, but a little bit more garbage.
The r's are not part of that rule, that's all im trying to say.

Answer 28

do you think I should do \[8^2\]? Or just write 8?

Answer 29

For what? 0_O

Answer 30

oh sorry, for complex conjugate if r=8 for the original

Answer 31

before the complex conjugate

Answer 32

So you're asking, what is the complex conjugate of this?\[\Large\rm 8(\cos \theta+ i \sin \theta)\]Well I guess I'm forgetting about the fact that you can distribute the 8,\[\Large\rm 8\cos \theta+8i \sin \theta\]So the complex conjugate would be,\[\Large\rm 8\cos \theta -8i \sin \theta=8(\cos \theta-i \sin \theta)\]

Answer 33

1.43 was tnat supposed to be root2 lol

Answer 34

@xapproachesinfinity yea

Answer 35

hehe that's weird in decimal and it is 1.41 not 1.43

Answer 36

Oh, wait, it isn't

Answer 37

it is just 1.43, not root 2

Answer 38

The problem gave the equation liek that

Answer 39

oh im not thinking that's angle in radians lol

Answer 40

oh kk

Answer 41

Thanks though

Answer 42

@zepdrix

If I need to turn (4(cos(7pi/9)+isin(7pi/9)))^3 into the form a+bi, would I just distribute the 3?

Answer 43

Distribute the 3 as in....
Expanding out three sets of brackets? No.
Or do you mean by apply De'Moivre's Rule? Yes.

Answer 44

as in multiplying theta by 3 and cubing 4

Answer 45

so de movire I think

Answer 46

ya that seems like the way to go :)

Answer 47

awesome :D Thanks for the help!

Answer 48

yay team

Answer 49

good work :)